Problem: Factor the following expression: $-9$ $x^2+$ $43$ $x+$ $10$
This expression is in the form ${A}x^2 + {B}x + {C}$ . You can factor it by grouping. First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-9)}{(10)} &=& -90 \\ {a} + {b} &=& & & {43} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-90$ and add them together. Remember, since $-90$ is negative, one of the factors must be negative. The factors that add up to ${43}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-2}$ and ${b}$ is ${45}$ $ \begin{eqnarray} {ab} &=& ({-2})({45}) &=& -90 \\ {a} + {b} &=& {-2} + {45} &=& 43 \end{eqnarray} $ Next, rewrite the expression as ${A}x^2 + {a}x + {b}x + {C}$ $ {-9}x^2 {-2}x +{45}x +{10} $ Group the terms so that there is a common factor in each group: $ ({-9}x^2 {-2}x) + ({45}x +{10}) $ Factor out the common factors: $ x(-9x - 2) - 5(-9x - 2) $ Notice how $(-9x - 2)$ has become a common factor. Factor this out to find the answer. $(-9x - 2)(x - 5)$